concatenate char with int

I have to concatenate char with int. Here's my code:

int count = 100;
char* name = NULL;
sprintf((char *)name, "test_%d", count);
print开发者_StackOverflow社区f("%s\n", name);

Nothing printed. What's the problem?

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You didn't allocate any memory into which sprintf could copy its result. You might try:

int count = 100;
char name[20];
sprintf(name, "test_%d", count);
printf("%s\n", name);

Or even:

int count = 100;
char *name = malloc(20);
sprintf(name, "test_%d", count);
printf("%s\n", name);

Of course, if your only goal is the print the combined string, you can just do this:

printf("test_%d\n", 100);

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If you programm C++ use sstream instead:

stringstream oss;
string str;
int count =100

oss << count;
str=oss.str();

cout << str;

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You have to allocate memory for name first. In C, library functions like sprintf won't make it for you.

In fact, I am very surprised that you didn't get a segmentation fault.

A simple workaround would be using char name[5+11+1] for the case of 32-bit int.

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I use boost::format for this.

#include <boost/format.hpp>

int count = 100;
std::string name = boost::str( boost::format("test_%1%") % count );

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Since the answer is tagged C++, this is probably how you should do it there:

The C++11 way: std::string str = "Hello " + std::to_string(5);

The Boost way: std::string str = "Hello " + boost::lexical_cast<std::string>(5);

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#include <iostream>
#include <string>
#include <sstream>  

int count = 100;

 std::stringstream ss;
 ss << "Helloworld";
 ss << " ";
 ss << count ;
 ss << std::endl;
 std::string str = ss.str();
 std::cout << str;

 const char * mystring = str.c_str();